jmc

The diagonals of a rhombus intersect at a 90-degree angle, partitioning the rhombus into four congruent right triangles. The legs of one of the triangles are 6 feet and 9 feet, so the hypotenuse of the triangle - which is also the side of the rhombus - is $\sqrt{(6^2+9^2)}=\sqrt{(36+81)}=\sqrt{117}$ feet. Since $117=9\times 13$, we can simplify this as follows: $\sqrt{117}=\sqrt{(9\times 13)}=\sqrt{9}\times \sqrt{13}=3\sqrt{13}$ feet. The perimeter of the rhombus is four times this amount or $4\times 3\sqrt{13}=\boxed{12\sqrt{13}\text{ feet}}$.